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AP Physics Notes

Table of Contents

2/3: Motion & Kinematics

This course works with only constant acceleration and, therefore, only linear or constant velocity.

One-Dimensional Motion

Velocity is equal to displacement (change in position) over time.

Acceleration is equal to change in velocity over time.

When an object moves with constant acceleration, the average acceleration equals the instantaneous acceleration. Therefore, $a=\frac{v_f-v_0}{t}$$a = \frac{v_f - v_0}{t}$, which can be rewritten as:

This makes sense, as with linear velocity, the final velocity over a given time period will equal the velocity at the starting point, plus the change in velocity ($\mathrm{\Delta }v$$\Delta v$), which can be derived from (2): $\mathrm{\Delta }v=at$$\Delta v = at$.

With linear velocity, average velocity is equal to displacement over time, but it is also equal to the average of the initial and final velocities.

Going back to (1), we can conclude that $v=\frac{\mathrm{\Delta }x}{t}=\frac{v_0+v_f}{2}$$v = \frac{\Delta x}{t} = \frac{v_0 + v_f}{2}$.

This can be rewritten as:

This intuitively makes sense, as when calculating for the area under underneath a linear velocity graph segment (which is displacement), this equation gives the formula for that area.

If we go back to (3), we can substitute in $v_f=v_0+at$$v_f = v_0 + at$ into (5), getting: $\mathrm{\Delta }x=\frac{1}{2}(v_0+v_0+at)t$$\Delta x = \frac{1}{2}(v_0 + v_0 + at)t$.

Simplified and standardized, this becomes:

This can also be derived using simple integration, which makes more practical sense:

We can also easily visualize it with the following standard gravity function that describes position as a function of time for a projectile:

Notice how this exactly describes (6).

The following equation can be derived to get an equation that does not use time.

This equation feels arbitrary because it is arbitrary. It doesn't make practical sense and it cannot be intuitively visualized. It is purely used to make calculations easier. It can, however, be completely avoided. The following problem will first be solved without the equation, then it will be solved with the equation:

A jet is taking off from the deck of an aircraft carrier. Starting from rest, the jet is catapulted with a constant acceleration of $+31\frac{\text{m}}{{\text{s}}^2}$$+31 \frac{\text{m}}{\text{s}^2}$ along a straight runway and reaches a velocity of $+62\frac{\text{m}}{\text{s}}$$+62 \frac{\text{m}}{\text{s}}$. Find the displacement of the jet.

Without the use of (8) (notice how this process involves solving for time; (8) skips this step):

With the use of (8):

Projectile Motion

• X Component

  $$\begin{array}{}\text{(9)}& \begin{array}{c}x(t)=v_0(\cos \theta )t\\ x(t)=v_{x_0}t\end{array}\end{array}$$

• Y Component

  $$\begin{array}{}\text{(10)}& \begin{array}{c}y(t)=\frac{1}{2}a{t}^2+v_0(\sin \theta )t\\ y(t)=\frac{1}{2}a{t}^2+v_{y_0}t\end{array}\end{array}$$

Using these two equations, the path of a projectile can be modeled neglecting air resistance, given the original direction and magnitude of the projectile. $v_0$$v_0$ is the original magnitude of the velocity, and $\theta$$\theta$ is the direction.

Usually, for projectile motion, the acceleration is gravity.

Relative Velocity

The velocity of $a$$a$ relative to $b$$b$ is denoted by using the vector $v_{ab}$$v_{ab}$.

To figure out the velocity relative to something else, all of the intermediate factors have to be eliminated.

Remember that these are vectors.

Another important thing to keep in mind is how to inverse the relative vectors.

4: The Laws of Motion

Types of Forces

These are the forces that always act upon everything in the universe.

• Gravitational Force - attractive force that exists between all objects

• Electromagnetic Force - forces that give materials their strength, their ability to bend and squeeze, stretch, or shatter

• Weak Force - a form of electromagnetic force involved in the radioactive decay of some nuclei

• Strong Nuclear Force - holds the particles in a nucleus together

All types of forces can be classified as one of the two following types.

• Contact Forces - forces that result from physical contact between two objects

• Field Forces - forces that do not involve physical contact between two objects

Newton's First Law of Motion

Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

• When the net external force on an object is 0, its acceleration is 0.

• Inertia is the tendency of an object to resist any attempt to change its motion.

• Mass is a quantitative measurement of inertia

• There is not very much difference between an object at rest and one in motion at constant velocity

Newton's Second Law of Motion

• Force is directly proportional to acceleration.

• Mass is inversely proportional to acceleration.

• Force is a vector quantity.

Newton's Third Law of Motion

Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object.

Free Body Diagrams & Sums of Forces

Free body diagrams display all forces acting on a single object. Sums of forces equations can then be written based off of free body diagrams to calculate certain values. The axis of free body diagrams can be titled in any way to ensure the most efficient solving process.

Sums of forces equations equal $0\text{N}$$0 \text{N}$ if the object is not accelerating and equal $ma$$ma$ if the object is accelerating, where $m$$m$ is the mass of that one single object alone.

Let $x_p$$x_p$ be all forces in the positive x direction. Let $x_n$$x_n$ be all forces in the negative x direction. Let $y_p$$y_p$ be all forces in the positive y direction. Let $y_n$$y_n$ be all forces in the negative y direction.

Friction

There are two types of friction:

• Static Friction

  • Represented by ${\overrightarrow{F}}_s$$\vec{F}_s$

  • Only applies when the object is not moving

  • If the external force (applied force) is greater than ${\overrightarrow{F}}_s$$\vec{F}_s$, only then does the object accelerate.

• Kinetic Friction

  • Represented by ${\overrightarrow{F}}_k$$\vec{F}_k$

  • The object only accelerates if this force is less than the external force; otherwise, it has constant velocity.

Kinetic friction is always less than static friction for the same surfaces and objects.

Both types of frictional forces are proportional to the normal force on the object.

The proportionality constant between friction and normal force is known as the coefficient of friction, and it is denoted by $\mu$$\mu$. It varies based off of materials in constant, properties of the materials, dryness of the materials, etc.

5: Work and Energy

Energy is the ability to do work.

Energy is never lost, only converted. (This is the law of conservation of energy.)

Work

The unit of work is the joule (J).

If $\theta =0$$\theta = 0$, because $F=ma$$F = ma$, the following can be derived:

Energy

Kinetic Energy:

Because kinetic energy is the same as the amount of work put into an object's motion if that object was originally at rest (this is due to the law of conservation of energy), we can take the above derived work equation, set $W=\mathit{KE}$$W = \mathit{KE}$ and $v_0=0$$v_0 = 0$, and get the following equation.

Another intuitive way to derive kinetic energy is using calculus as follows. It is important to note that $\int F\mathit{dx}=W$$\int F \mathit{dx} = W$¹.

Work-Energy Theorem:

Kinetic energy is the work an object is capable of doing in coming to rest.

Potential Energy:

Potential energy is the energy an object has as a result of its position. Because of the law of conservation of energy, we can take the equation $W=max$$W = max$ and set $W=\mathit{PE}$$W = \mathit{PE}$. Because potential energy is the energy in an object due to position, $a$$a$ is going to be gravity. We can therefore replace $a$$a$ with $g$$g$. Finally, for practical purposes, the distance, $x$$x$, is always going to be the height of the object, so, for easy referencing, we rename $x$$x$ to $h$$h$ to get the following equation.

There is an intuitive way of thinking about the potential energy in an object. Imagine raising a ball up to $2\text{m}$$2\text{m}$. Let's say you put in $x$$x$ amount of work to accomplish this. Where does that work go? Is the energy you put in lost? No (law of conservation of energy), it's not. That energy is now stored in the ball as potential energy. Now imagine letting go of the ball. On its way down, the ball releases the same amount of energy you put into it. While the ball was up at a height of $2\text{m}$$2\text{m}$, this energy was stored in the ball as potential energy.

Potential energy is always relative.

Energy is never lost, only converted:

Spring/Elastic Potential Energy

Energy stored in a compressed spring is called spring or elastic potential energy.

To compress a spring, you must exert a force $F$$F$ to compress it $x$$x$ amount. The ratio of these two amounts is known as the spring constant of a given spring, denoted by $k$$k$. When pushing on a spring, the spring must push back with the same amount of force. Therefore, we can derive the following. $F_s$$F_s$ is the force that the spring exerts when compressed. There is a negative, because the direction that the spring pushes or pulls is in the opposite direction as displacement.

Elastic potential energy is defined as the amount of work a spring is capable of doing. Remember that $W=Fx$$W = Fx$. However, in the case of a spring, the force is dependent on the distance. $F$$F$ is a function of $x$$x$, so to calculate the total capable work, we can take the following integral (we can omit the negative from $F=-kx$$F = -kx$ because the direction does not matter when considering the work being done):

A revised version of (22) would therefore be:

All this is saying is that all types of energy combined initially is the same as all types of energy combined finally due to the law of conservation of energy.

Including the work done by non-conservative forces which add or take away energy², we can conclude the following ($W_{nc}$$W_{nc}$ is the work done by non-conservative forces):

A spring-related concept that tends to be confusing is the comparison of the work done by gravity and a spring. For example, take the following problem:

A spring of spring constant $k=8.75\frac{\text{N}}{\text{m}}$$k = 8.75 \frac{\text{N}}{\text{m}}$ is hung vertically from a rigid support. A mass of $0.500\text{kg}$$0.500 \text{kg}$ is placed on the end of the spring and supported by hand at a point so that the displacement of the spring is $0.250\text{m}$$0.250 \text{m}$. The mass is suddenly released and allowed to fall. At the lowest point of the mass, what is the displacement of the spring from its equilibrium position?

The difficult concept to relate here is: Why does the mass come to a stop when the work done by gravity and the spring are equal?

A common misconception is to assume that the mass would come to rest once the force pulling up and pulling down is the same. With equal forces, however, it is only guaranteed that the acceleration is zero, not the velocity. Therefore, we have to relate something else.

The mass starts at rest and ends at rest, which means that $\mathrm{\Delta }\mathit{KE}=0$$\Delta \mathit{KE} = 0$. The change in kinetic energy, though, is just the work done by the spring plus the work done by gravity. So: $\mathrm{\Delta }\mathit{KE}=W_g+W_s=0$$\Delta \mathit{KE} = W_g + W_s = 0$. So: $W_g=-W_s$$W_g = -W_s$. Because they pull in opposing directions, depending on how you establish your coordinate system, either $W_g$$W_g$ or $W_s$$W_s$ will be negative, so: $W_g=W_s$$W_g = W_s$.

For a more sophisticated explanation, visit this physics stack exchange page.

Power

Power is the rate at which work is applied. Therefore, the derivative of work with respect to time is power.

Therefore, average power is the average work (or constant work) over change in time.

The units of power is $\frac{\text{J}}{\text{s}}$$\frac{\text{J}}{\text{s}}$ which is known as the watt ($\text{W}$$\text{W}$), not to be confused with $W$$W$ (work).

We can expand (28) to get slightly different forms of the equation:

(30) is the instantaneous power exerted by a force exactly at the time when velocity is $v$$v$.

6: Momentum & Collisions

Momentum

Momentum is inertia in motion.

The units of momentum are $\frac{\text{kg}\text{m}}{\text{s}}$$\frac{\text{kg}\text{m}}{\text{s}}$ or $\text{N}\text{s}$$\text{N}\text{s}$.

Impulse

Taking Newton's second law, $F=ma$$F=ma$, we can substitute $a$$a$ for $\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$\frac{\Delta v}{\Delta t}$, getting $F=\frac{m\mathrm{\Delta }v}{\mathrm{\Delta }t}$$F=\frac{m \Delta v}{\Delta t}$. Multiplying both sides of the equation by $\mathrm{\Delta }t$$\Delta t$, we can get:

And also:

This pretty much means that mass times the change in velocity is the change in momentum. The change in momentum is called impulse. $F\mathrm{\Delta }t$$F \Delta t$ is also, therefore, impulse. Using all three of these, we can get the following impulse equation:

Impulse:

Impulse can also be defined as a force applied over a certain amount of time. In other words, a force applied to an object for a certain time causes its momentum to change proportionally to the force applied and the time it was applied for.

Conservation of Momentum

Given any collision, the momentum before always equals the momentum after. Momentum is always conserved. Kinetic energy, however, is not always conserved. The conservation of kinetic energy is dependent on the types of collisions...

Types of Collisions

• Perfectly Inelastic Collisions

  • 2 objects get stuck together

  • Momentum is conserved but kinetic energy is not

  • Example: Dropping a sticky ball that does not bounce at all

• (Partially) Inelastic Collisions

  • 2 objects bounce off each other

  • Momentum is conserved and so is some kinetic energy, but not all kinetic energy

  • Example: Dropping a ball that bounces back up, but not to the same height you dropped it from

• Elastic Collisions

  • 2 objects bounce off each other

  • Momentum is conserved and so is kinetic energy

  • Example: Dropping a ball that bounces back up to the same exact height you dropped it from

Most real world collisions fall in the category of partially inelastic collisions.

In elastic collisions, it is important to consider that kinetic energy is completely conserved. This fact allows us to solve problems that would otherwise not be solvable. The following equation can be derived:

All this equation is saying is that the combined velocity of the first object is equal to the combined velocity of the second object.

To envision this, imagine a Newton's Cradle apparatus. Think about the velocities of the 2 spheres on the end before and after the collision. It lines up with (35).

Glancing Collisions

Glancing collisions works with higher dimensions of momentum vectors. This applies when an object hits another one at an angle, and the resulting velocities of both objects are also at different angles.

The general idea is that glancing collisions involve 2-dimensional vectors instead of 1-dimensional vectors.

It is very useful to draw a picture for these types of problems. There are 2 ways to do these types of problems:

1. Separate the $x$$x$ axis and $y$$y$ axis into their own separate equations using the components of the momentums

2. Represent all momentum quantities as vectors throughout the solving process

Here is an example problem:

At an intersection, a $1500\text{kg}$$1500 \text{kg}$ car traveling east at $25\frac{\text{m}}{\text{s}}$$25 \frac{\text{m}}{\text{s}}$ collides with a $2500\text{kg}$$2500 \text{kg}$ van traveling north at $20\frac{\text{m}}{\text{s}}$$20 \frac{\text{m}}{\text{s}}$. Find the direction and magnitude of the velocity of the wreckage immediately after the collision, assuming that the vehicles undergo a perfectly inelastic collision.

Solving using strategy 1:

Solving using strategy 2:

Both of the methods work. Which one to use depends on personal preference.

7: Circular Motion and the Law of Gravity

This portion of the course works with radians. Radians have no units, as they are a ratio between arc length and radius.

When rotating a line segment along the origin of a circle, once the distance covered by the end of the line segment is equal to the radius, the angle rotated is one radian.

Using this definition, we can say that:

Remember, radians have no units. Using the above conversion factor, we can convert radians to degrees and vice-versa whenever needed.

Angular Displacement

Angular displacement is defined as the difference between the original angle and final angle (change in angle) in radians. Angular displacement is $\mathrm{\Delta }\theta$$\Delta \theta$.

Angular Velocity

Angular velocity is denoted by $\omega$$\omega$. It is defined by angular displacement over change in time.

The units of angular velocity are ${\text{s}}^{-1}$$\text{s}^{-1}$ because radians don't have units.

Angular Acceleration

Angular acceleration is denoted by $\alpha$$\alpha$. It is defined by change in angular velocity over change is time.

The units of angular acceleration are ${\text{s}}^{-2}$$\text{s}^{-2}$.

Relations Between Angular & Linear Quantities

Arc length is usually denoted by $s$$s$.

Going back to the definition of a radian, we can say the angle in radians is equal to arc length over radius because radians are the ratio of arc length and radius. Therefore:

This relates linear displacement and angular displacement. Our next goal is to relate linear velocity to angular velocity. To do this we need to derive $\frac{\mathrm{\Delta }\theta }{\mathrm{\Delta }t}$$\frac{\Delta \theta}{\Delta t}$ because $\omega =\frac{\mathrm{\Delta }\theta }{\mathrm{\Delta }t}$$\omega = \frac{\Delta \theta}{\Delta t}$. Also, note that $\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}=v_t$$\frac{\Delta s}{\Delta t} = v_t$. $v_t$$v_t$ is the tangential velocity. Tangential velocity is the instantaneous velocity which, with angular motion, happens to be tangential.³

We can use (39) to arrive here as follows:

The next goal is to relate linear acceleration to angular acceleration. To do this we need to derive $\frac{\mathrm{\Delta }\omega }{\mathrm{\Delta }t}$$\frac{\Delta \omega}{\Delta t}$ using the same method as in the derivation of (40).

All of these derivation have a direct correlation with each other considering the radius.

Here are all the equations listed:

Centripetal Acceleration/Centripetal Force

Contrary to popular belief, centripetal force doesn't push outward, it pushes inward.⁴ The only reason people think it pushes outward is actually because of inertia. The tendency for an object to not move wants to make it leave angular motion, in order for it to stay in angular motion, there must be a force pushing back or an acceleration toward the center of the circle. This is called centripetal acceleration.

A really important concept to understand is that centripetal acceleration/centripetal force is not some external force that just pulls on an object. It is the resulting force of other forces. For example, when a car is traveling in a circle, the friction in the turned wheels causes there to be a force that pushes inward. This is called centripetal force. There isn't just some external force acting on the car making it go in a circle. If the wheels were straight, the frictional force to provide the centripetal force wouldn't be there, and the car would go straight. Similarly, on a banked curve, the $x$$x$ component of the normal force is what provides the centripetal force.

The derivation of centripetal acceleration is a complex one. However, I will try my best to explain. centripetal acceleration is denoted by $a_c$$a_c$.

Look at the diagram below for reference on the derivation.

Given a change in time, $\mathrm{\Delta }t$$\Delta t$, the object in question travels from point $A$$A$ to point $B$$B$.

The magnitude of ${\overrightarrow{v}}_i$$\vec v_i$ and ${\overrightarrow{v}}_f$$\vec v_f$ is the same. Let's call the magnitude of both velocities just $v$$v$.

As seen on the diagram, $\mathrm{\Delta }\overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i$$\Delta \vec v = \vec v_f - \vec v_i$.

The acceleration we're interested in is $\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$\frac{\Delta v}{\Delta t}$. Notice how the numerator is $\mathrm{\Delta }v$$\Delta v$ and not $\mathrm{\Delta }\overrightarrow{v}$$\Delta \vec v$. This is because we are only interested in the magnitude, as the centripetal acceleration always points toward the center of the circle.

It is really important to note that the acceleration we're interested in is the instantaneous acceleration, therefore $a_c=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$$a_c = \lim\limits_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}$. Also, as $\mathrm{\Delta }t\to 0$$\Delta t \rightarrow 0$, $\theta \to 0$$\theta \rightarrow 0$.

The triangle created by $\mathrm{\Delta }\overrightarrow{v}$$\Delta \vec v$, ${\overrightarrow{v}}_f$$\vec v_f$, and ${\overrightarrow{v}}_i$$\vec v_i$ is similar to the triangle created by both $r$$r$'s and $\overline{AB}$$\overline {AB}$ as $\theta \to 0$$\theta \rightarrow 0$. Because they are similar, in the diagram, we can see $\theta$$\theta$ in 2 places.

As $\theta \to 0$$\theta \rightarrow 0$, the arc lengths can be considered straight. Therefore, $\mathrm{\Delta }v$$\Delta v$ can be considered equal to the arc length of $\theta$$\theta$ and $v$$v$. Going back to the arc length formula, $\mathrm{\Delta }v=v\theta$$\Delta v = v \theta$ because $s=r\theta$$s = r \theta$.

We now have $\mathrm{\Delta }v$$\Delta v$, we now need $\mathrm{\Delta }t$$\Delta t$ to get acceleration.

$\theta =\omega \mathrm{\Delta }t$$\theta = \omega \Delta t$ because $\omega =\frac{\theta }{\mathrm{\Delta }t}$$\omega = \frac{\theta}{\Delta t}$, so $\mathrm{\Delta }t=\frac{\theta }{\omega }$$\Delta t = \frac{\theta}{\omega}$.

Since $v_t=r\omega$$v_t = r \omega$, $\omega =\frac{v_t}{r}$$\omega = \frac{v_t}{r}$.

In our case, $v_t=v$$v_t = v$, so $\omega =\frac{v}{r}$$\omega = \frac{v}{r}$. Plugging this into the $\mathrm{\Delta }t$$\Delta t$ equation, we get that $\mathrm{\Delta }t=\frac{\theta }{\frac{v}{r}}=\frac{r\theta }{v}$$\Delta t = \frac{\theta}{\frac{v}{r}} = \frac{r \theta}{v}$.

We now have $\mathrm{\Delta }v$$\Delta v$ and $\mathrm{\Delta }t$$\Delta t$, so we can get acceleration.

Here's the purely mathematical way to derive it:

This YouTube video explains how centripetal force works really well.

Centripetal force is is just centripetal acceleration times mass:

Universal Law of Gravitation

Every particle in the universe is attracted to every other particle in the universe. The force of attraction between them is called gravity. The force of gravity is directly proportional to the product of the mass of two object and inversely proportional to the square of the distance between their center of masses:

It might be useful to ask where this equation comes from. Unfortunately, though, this equation doesn't have a derivation.⁵

$G$$G$ is the proportionality constant, known as the constant of universal gravitation or the universal gravitational constant.

Gravitational Potential Energy

With objects near Earth's surface, the equation ${\mathit{PE}}_g=mgh$$\mathit{PE}_g = mgh$ works. However, as we move farther away from Earth by significant amounts, gravity starts to lessen, and, in turn, gravitational potential energy changes. We must redefine gravitational potential energy.

The previous equation only described the change in gravitational potential energy. This new definition will get us the total value of gravitational potential energy, which can be defined as: the work needed to bring an object from infinitely far away (where gravity is $0$$0$ and GPE is $0$$0$) to a distance $r$$r$ away.

Because gravity naturally brings objects close together, the work required by an external force to bring objects together is negative because it happens naturally.

The force of gravity gets exponentially less as distance increases. The work done by gravity is force multiplied by distance. Remember that $W=\int F\mathit{dx}$$W = \int F \mathit{dx}$. With a constant force, this evaluates to $W=Fx$$W = Fx$. However, with gravity, the force changes with respect to distance. Our distance variable in this case is $r$$r$, and the formula for force of gravity is (45). Putting everything together, we can get the following equation:

The bounds of the integral are from $\mathrm{\infty }$$\infty$ to $r$$r$ because GPE is

the work gravity is capable of doing in bringing an object from infinitely far away (Where gravity is 0 and GPE is 0) to a distance $r$$r$ away.

Solving the integral in (47), we get:

At $r=\mathrm{\infty }$$r=\infty$ the force of gravity and gravitational potential energy is $0$$0$, so the second half of the equation is $0$$0$.

Therefore, the formula for total GPE is:

This YouTube video explains the derivation really well.

Escape Velocity

Escape velocity is the lowest possible velocity an object can have in order to escape the gravitational pull of another object.

There are a few things we must note to be able to derive the formula for escape velocity. The total energy of an object is the kinetic energy plus the gravitational potential energy.

As an object moves away, gravity will lessen its velocity, so the kinetic energy will eventually turn into gravitational potential energy. The escape velocity describes the velocity required to completely escape the gravitational pull of an object. In theory, this can only be achieved at an infinite distance.

Therefore, for theoretical purposes, we can assume that the object is capable to moving infinitely far away. At this distance, gravitational potential energy is $0$$0$ and all kinetic energy will have been converted to gravitational potential energy, so $\mathit{KE}+{\mathit{PE}}_g=0$$\mathit{KE} + \mathit{PE}_g = 0$. We can now solve for the escape velocity. Let $m_e$$m_e$ be the mass of the object that is escaping and $m$$m$ be the mass of the object being escaped from.

Solving for $v_e$$v_e$, we get:

Notice how escape velocity doesn't depend on the escaping object's mass, only the mass of the object it is escaping from.

This YouTube video explains the derivation.

Kepler's Laws

1. All planets move in elliptical orbits with the Sun at one of the focal points.

2. A line drawn from the Sun to any planet sweeps out equal area in equal time intervals.

   All this means is that a planet moves faster when it is closer to the Sun but slower when it's farther away.

3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun.

   All this means is that the orbital period for every planet is dependent on how far away (on average) it is from the Sun.

   This can be derived in equation form.

   Going back to centripetal force and acceleration, the sums of forces equation for an orbiting planet would be ($m_s$$m_s$ is the mass of the sun, $m_p$$m_p$ is the mass of the planet and $T$$T$ is the orbital period):⁶

   $$\begin{array}{rl}& F_c=m_p{a}_c\\ & G\frac{m_s{m}_p}{r^2}=m_p\frac{v^2}{r}\\ & v=\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}=\frac{2\pi r}{T}\\ & \frac{G{m}_s}{r^2}=\frac{1}{r}(\frac{2\pi r}{T}{)}^2\\ & \frac{G{m}_s}{r^2}=\frac{4{\pi }^{2}r}{T^2}\end{array}$$

   Rearranging, we get:

   $$\begin{array}{}\text{(52)}& T^2=(\frac{4{\pi }^2}{G{m}_s})r^3\end{array}$$

   The proportionality constant can be denoted by $K_s$$K_s$.

   $$\begin{array}{}\text{(53)}& T^2=K_s{r}^3\end{array}$$

   $K_s$$K_s$ can be defined as the following:

   $$\begin{array}{}\text{(54)}& K_s=\frac{4{\pi }^2}{G{m}_s}\end{array}$$

8: Rotational Equilibrium & Rotational Mechanics

Torque

Torque is the ability of a force to rotate a body about some axis. It is measured as the product of the force being applied perpendicularly to the axis of rotation. Torque is denoted by $\tau$$\tau$.

$F$$F$ is the force and $r$$r$ is the perpendicular radius from the axis of ration to the force.

The units of torque are $\text{Nm}$$\text{Nm}$. Do not put $\text{J}$$\text{J}$ as it is highly unconventional.

The positive/negative signs of torque are consistent with the signs of radians. Moving up from the positive $x$$x$ axis (counterclockwise) is positive torque, and moving down from the positive $x$$x$ axis (clockwise) is negative torque.

Sums of torque diagrams and equations can be created in the same manner as sums of forces diagrams and equations to solve for unknowns. In rotational equilibrium, the sums of torque is equal to $0\text{Nm}$$0 \text{Nm}$. If an object is in rotational equilibrium, it does not matter where you put the axis of rotation.

An intuitive way to think about torque is imagining opening a door. Pushing closer to the hinges requires more force to achieve the same rotation as pushing farther away from the hinges. Therefore, if the force was the same, you would exert less torque when pushing closer to the hinges because the door wouldn't open as much. If you applied the same force to the edge of the door, the door would move a lot more, so the torque exerted would be greater.

Another way to think of torque is the strength/force of rotation.

Center of Gravity/Center of Mass

When referring to objects with a weight, we usually consider the weight to be concentrated at one point in space. This is called the center of gravity or center of mass.

To calculate where the center of mass is, we have to compute a sort-of average on all the points of mass in a system. To start, pick a point that every other point will be relative to.

The center of mass relative to that point in the $x$$x$ direction will then be (Let $n$$n$ be the number of points of mass in the system):

Similarly, the center of mass relative to that point in the $y$$y$ direction would be:

Rotational Kinematics

Because of the similarities between linear and rotational quantities, linear kinematic equations hold true for rotational kinematics.

Relationship Between Torque and Angular Acceleration

When the net torque on an object is not $0\text{Nm}$$0\text{Nm}$, it undergoes angular acceleration.

Torque is equal to $Fr$$Fr$, and $F=ma$$F = ma$. If we try to get a derivation for torque by multiplying both sides of the force equation by $r$$r$, we get:

Moment of Inertia

The moment of inertia is generally thought of as the inertia of rotating objects. For example, a rotating circular object that has more mass located farther away from its origin is harder to start rotating and stop rotating than an object whose mass is closer to the origin. This is measured as the moment of inertia.

Notice the similarities between the two equations:

In the linear equation, the mass is what provides the inertia, but in the angular equation, the moment of inertia is provided by $m{r}^2$$mr^2$. That above equation only works, though, if the rotating object only has mass concentrated at one point which is a distance $r$$r$ away from the origin.

In order to get an equation to accurately model real world objects where there is mass spread out, we have to sum up every point's moment of inertia. This is denoted by $I$$I$.

Even though $I$$I$ and $m$$m$ have similarities, mass is an intrinsic property of an object that does not change, whereas the $I$$I$ of a system depends upon the axis of rotation and the manner in which the mass is distributed.

The full relationship is:

Moments of Inertia of Extended Objects

To calculate moments of inertia for extended objects, you need integral calculus. For this reason, all the formulas of moments of inertia of extended objects will be given on tests etc.

The following part of this subsection is NOT needed for this course.

If you are curious, though, as to how to actually calculate these moments of inertia, I will walk you through it.

Given a finite number of points with finite masses that make up an object we know that the moment of inertia is equal to the sum of them:

Now, if we take an extended object, each point will be infinitesimally small and would, therefore, have an infinitesimally small mass. The radius, however, will stay the same.

We cannot directly relate $r$$r$ and $dm$$dm$, but we cannot pull $r^2$$r^2$ out of the integral because the two variables are dependent on each other.

For one example, let's calculate the moment of inertia of a rod of length $L$$L$ and mass $m$$m$ that rotates about the center.

Because the rod rotates about the center and its length is $L$$L$, we can get the following limits of integration:

To relate mass to linear quantities, we can use the linear mass density of the object, denoted by $\lambda$$\lambda$, which is the mass per unit length. Therefore:

Differentiating both sides and keeping in mind that the linear mass density is a constant, we get:

$dl$$dl$ represents an infinitesimal piece of $r$$r$, which means that $dl=dr$$dl = dr$. Putting everything together:

Evaluating the integral, we obtain:

Remember that $\lambda =\frac{m}{l}$$\lambda = \frac{m}{l}$, and also in this case $l=L$$l = L$.

This was the derivation of the moment of inertia of a rod of length $L$$L$ and mass $m$$m$ that rotates about the center.

Other objects' moment of inertia can be derived in very similar ways. This was just one example.

Rotational Kinetic Energy

Remember the original kinetic energy formula, $\mathit{KE}=\frac{1}{2}m{v}^2$$\mathit{KE} = \frac12 mv^2$. As we've seen before with moment of inertia, for rotating objects, we must consider the sum of all points with mass. Let the rotational kinetic energy of an object be the sum of the kinetic energy of all points with mass $m$$m$ and tangential velocity $v$$v$.

Converting tangential velocity to angular velocity using $v=r\omega$$v = r\omega$, we get:

The sum of angular velocities is just $\omega$$\omega$ and can even be considered a constant, so we can take it out of the sum.

Remember the equation for moment of inertia, $I=\sum m{r}^2$$I = \sum mr^2$. Substituting this in, we get:

Comparing this to the original kinetic energy equation, this should not be surprising considering the similarities we've seen in the past. Going from translational to rotational, $I$$I$ usually takes the place of $m$$m$ and $\omega$$\omega$ usually takes the place of $v$$v$.

Angular Momentum

Keep in mind the similarities between $m$$m$ and $I$$I$, and $v$$v$ and $\omega$$\omega$. Angular momentum is denoted by $L$$L$.

Because translational momentum is $p=mv$$p = mv$, angular momentum is therefore (considering the similarities):

Unlike translational moment, whose units are $\frac{\text{kgm}}{\text{s}}$$\frac{\text{kgm}}{\text{s}}$, the units of angular momentum are $\frac{{\text{kgm}}^2}{\text{s}}$$\frac{\text{kgm}^2}{\text{s}}$.

Angular momentum also means there is rotational impulse. It works very similarly to translational impulse.

Replacing $\alpha$$\alpha$ with $\frac{\mathrm{\Delta }\omega }{\mathrm{\Delta }t}$$\frac{\Delta \omega}{\Delta t}$, we obtain $I\mathrm{\Delta }\omega$$I \Delta \omega$ in the numerator, which just represents change in $L$$L$ because:

Putting everything together as seen in (78), we obtain:

This is really similar to translational impulse. All this is saying is that, given an applied torque over a period of time, the angular momentum will change accordingly.

Conservation of momentum laws apply. (YouTube video example)

For single point particles with a known tangential velocity and radius, the angular momentum can be found using the following formula:

This is because $L=I\omega$$L = I \omega$, and, for single point particles, $I=m{r}^2$$I = mr^2$. Converting $\omega =\frac{v}{r}$$\omega = \frac{v}r$ and putting everything together: $L=I\omega =m{r}^2\frac{v}{r}=mvr$$L = I\omega = mr^2\frac{v}r = mvr$.

Vector Nature of Angular Quantities

Until now, we've considered only the magnitudes of angular quantities like $\omega$$\omega$, $\alpha$$\alpha$, and $L$$L$. For generalization purposes that don't depend on point of view, they can be considered vectors. The direction of the vector points along the axis of rotation.⁷

The direction of this vector can be found using the right hand rule. For this, you must use your right hand. Stick your thumb out and curl the rest of your finger. The direction that the rest of your fingers point should match the direction the object is rotating. The direction your thumb points is the direction of the vector. This works for angular velocity, angular acceleration, and angular momentum

9: Solids & Fluids

States of Matter

• Solids

  • Definite shape and definite volume

  • Can be crystalline (atoms with an ordered structure) or amorphous (random atomic structure)

  • Generally considered not compressible

• Liquids

  • Definite volume but indefinite shape

    • Intermolecular forces are not strong enough to keep the molecules in fixed positions

  • Generally considered not compressible

• Gases

  • Indefinite shape and indefinite volume

  • Gases are compressible

  • Gases fill up the container that they are in completely

Density

Density is defined as mass per volume. The more mass something has given the same volume, the more dense it is. Density is denoted by $\rho$$\rho$.

The units of density are $\frac{\text{kg}}{{\text{m}}^3}$$\frac{\text{kg}}{\text{m}^3}$.

The density of water is known to be $1\frac{\text{g}}{{\text{cm}}^3}$$1\frac{\text g}{\text{cm}^3}$. This is not in SI unit form. Converting this to SI unit form, we obtain the density of water to be $1000\frac{\text{kg}}{{\text{m}}^3}$$1000\frac{\text{kg}}{\text m^3}$.

The specific gravity of a substance is the ratio of its density to the density of water at $4^{\circ }\text{C}$$4^\circ \text C$ (This is just exactly where the density of water is exactly $1000\frac{\text{kg}}{{\text{m}}^3}$$1000\frac{\text{kg}}{\text m^3}$). For example, if the specific gravity of a substance is $3.7$$3.7$, its density is $3700\frac{\text{kg}}{{\text{m}}^3}$$3700\frac{\text{kg}}{\text m^3}$.

Pressure

Pressure is the force that's exerted over an area. The more concentrated the force (so the smaller the area it's spread out over), the greater the pressure. Pressure is denoted by $P$$P$.

The units of pressure are $\frac{\text{N}}{{\text{m}}^2}$$\frac{\text N}{\text m^2}$ which is equivalent to $\text{Pa}$$\text{Pa}$ (pascals).

Variation of Pressure with Depth

Within a fluid, all points at the same depth must have the same pressure as long as the fluid is in static equilibrium (it's not moving).

Look at the following figure:

The force pushing on top is $F_1$$F_1$ and the force pushing on the bottom is $F_2$$F_2$. The cylindrical cutout in the center is just a representation of some portion of the fluid within the bigger container.

Creating a sums of forces equation for the cutout portion of fluid using the free body diagram, we get:

It equals $0\text{N}$$0\text N$ because the fluid is in static equilibrium. Going back to the pressure equation, we know that $F=PA$$F = PA$. Therefore, we can write:

Going back again to the density equation, we know that $m=\rho V$$m = \rho V$. Volume, however, is just area times height, so $m=\rho Ah$$m = \rho Ah$. Substituting this in, we get:

Dividing each term by $A$$A$ and isolating $P_2$$P_2$, we finally get:

Look back at the diagram. This equation states that the pressure between the top and the bottom of a fluid, separated by a height $h$$h$ is the pressure at the top plus the density times $gh$$gh$. If the top of the fluid is exposed to the atmosphere, then the pressure at the top is $1\text{atm}$$1 \text{atm}$. $P_2$$P_2$ is always greater than $P_1$$P_1$.

The shape of the fluid container does not matter for this principle.

Pascal's Principle states that pressure applied to an enclosed fluid transmitted undiminished to every point of the fluid and to the walls of the containing vessel. All this states is that pressure always pushes evenly outward in all directions.

Hydraulics

In hydraulics, a U-shaped tube (called a u-tube) is used to be able to apply larger forces when a smaller force is initially put in.

Remember, the pressure at any same depth of a liquid is always the same, no matter the shape of the container.

Pressure Measurements

Look at the following diagram.

Our goal is the measure the pressure of the supplied gas.

$P_A$$P_A$ is the pressure of the gas and $P_0$$P_0$ is the atmospheric pressure. We know that $P_A>P_0$$P_A > P_0$ because of how the liquid is displaced.

Because the pressure at any same depth of a liquid is always the same, we know that the pressure at point $A$$A$ is the same as the pressure at point $C$$C$. Therefore, the pressure at point $C$$C$ is $P_A$$P_A$.

Using the variation of pressure with depth equation, we know that the pressure at the bottom of the displaced liquid is $P_0+\rho gh$$P_0 + \rho gh$.